NAME: PUID: ECE 305 Fall 017 Final Exam (Exam 5) Wednesday, December 13, 017 This is a closed book exam. You may use a calculator and the formula sheet at the end of this exam. Following the ECE policy, the calculator must be a Texas Instruments TI-30X IIS scientific calculator. To receive full credit, you must show your work (scratch paper is attached). The exam is designed to be taken 60 minutes (or less). However, the full two hours will be allowed, if you need them. Be sure to fill in your name and Purdue student ID at the top of the page. DO NOT open the exam until told to do so, and stop working immediately when time is called. The last pages are equation sheets, which you may remove, if you want. 100 points possible, I) 40 points (8 points per question) II) 30 points III) 30 points Course policy ---------------------------------- -- -- -- -- -- -- -------- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- If I am caught cheating, I will earn an F in the course & be reported to the Dean of Students. I repeat: Signature:
Part I: Answer the 5 multiple choice questions below by entering them on your IDP-15 Scantron. 1 (8 points). Which of the following factors directly increases the common emitter gain DC in a bipolar junction transistor (BJT)? a. Decreasing the collector doping b. Decreasing the emitter doping c. Increasing the emitter doping d. Increasing the base doping e. Increasing the collector doping (8 points). What is the greatest advantage of an npn heterojunction bipolar transistor (HBT), compared to a BJT? a. Electron current through the emitter is larger b. Hole current through the emitter is larger c. Electron current through the emitter is smaller d. Hole current through the emitter is smaller e. Electron current through the collector is larger 3 (8 points). Which of the label numbers in the diagram below on the right below corresponds to the operating regime depicted in the diagram below on the left? + - + - a. 1 b. c. 4 d. 5 e. 6
4 (8 points). Which of the following would decrease the magnitude of the Early effect in a BJT? a. Decreasing the collector doping b. Decreasing the emitter doping c. Increasing the emitter doping d. Decreasing the base doping e. Increasing the collector doping 5 (8 points). Which of the following components of the Ebers-Moll model can be neglected in the forward active mode, depicted below? a. 1, 4 b. 1, 3 c. 1 d., 4 e., 3
Part II (Free Response, 30 points) Consider an npn BJT made of silicon, depicted below. Assume that I Ep = 15 A, I En =.5 ma, I Cp = 130 na, and I Cn =.38 ma. Assume that recombination within the BJT is small (i.e., diffusion lengths are much larger than the thicknesses of each layer). a. What is the base transport factor α T? b. What is the emitter injection efficiency γ F? c. What are the total emitter, collector, and base currents I E, I C, and I B? d. What is the base current gain α dc and emitter current gain β dc? e. What would be the new value of I Ef the base region silicon were replaced by a hypothetical material with a bandgap of only 0.9 ev, but otherwise identical properties?
Part III (Free Response, 30 points) Consider adeal silicon pnp BJT at room temperature. The doping concentrations at emitter, base and collector are respectively N A,E = 10 17 cm -3, N D,B = 10 15 cm -3, and N A,C = 10 14 cm -3. The base width is W B. Assume that the minority carrier diffusion lengths L n,e = 40 W B, L p,b = 5 W B, and L n,c = 00 W B. The total and quasi-neutral base widths depicted below are schematic, and may not be to scale. a. Write an expression for the quasi-neutral base width W B as a function of W B, V EB, & V CB. You may use the one-sided junction approximation.
b. Sketch the excess minority carrier concentration across the emitter, base, and collector quasi-neutral regions, if V EB=0.5 V and V CB=0.3 V. Please indicate any assumptions about the emitter and collector thicknesses. What is the operating regime of the BJT under this condition? c. Given that the emitter injection efficiency F is 0.99, find the emitter current gain β dc. Assume that W B = 0.8 W B in this operating condition.
ECE 305 Exam 5 Formula Sheet (Fall 017) You may remove these pages from the exam packet, and take them with you. Physical Constants Silicon parameters (T = 300 K) h/π = ħ = 1.055 10 34 J s N C = 3.3 10 19 cm 3 m 0 = 9.109 10 31 kg N V = 1.83 10 19 cm 3 k B = 1.38 10 3 J/K = 1.1 10 10 cm 3 q = 1.60 10 19 C K s = 11.8 ε 0 = 8.854 10 1 F/m E g = 1.1 ev; χ = 4.03 ev Miller Indices: (hkl) {hkl} [hkl] <hkl> Density of states g C (E) = (m n ) 3/ (E E C ) π ħ 3 Fermi function f(e) = 1 1+e (E E F )/kt Intrinsic carrier concentration = N C N V e E g/kt Equilibrium carrier densities: N C = 1 4 (m n kt πħ )3/ N V = 1 4 (m p kt )3/ πħ n 0 = N C e (E F E C )/kt = e (E F E i )/kt p 0 = N V e (E V E F )/kt = e (E i E F )/kt Space charge neutrality: p n + N D + N A = 0 Law of Mass Action: n 0 p 0 = Non-equilibrium carriers: n = N C e (F N E C )/kt p = N V e (E V F P )/kt np = e (F N F P )/kt Conductivity/resistivity: σ = σ n + σ n = q(nμ n + pμ p ) = 1/ρ Drift-diffusion current equations: Carrier conservation equations: Poisson s equation: dn J n = nqμ n ℇ x + qd n = nμ df n dx n dx dp J p = pqμ p ℇ x qd p = pμ df p dx p dx n t = + (J n q ) + G n R n p t = (J p q ) + G p R p (εℇ) = ρ D n μ n = kt q D p μ p = kt q SRH carrier recombination: R = n/τ n or R = p/τ p Minority carrier diffusion equation: n = D n t n x n τ n + G L L D,n = D n τ n PN homojunction electrostatics: V bi = kt q ln (N DN A ) dℇ = ρ(x) dx K s ε o W = K sε o V bi ( N A+N D ) x q N A N n = ( N A ) W x D N A +N p = ( N D ) W D N A +N D ℇ(0) = qv bi ( N AN D ) K s ε o N A +N D
PN diode current: n(0) = N A (e qv A/kT 1) p(0) = n i (e qva/kt 1) N D J D = J o (e qv A/kT 1) J o = q ( D n L n N A + D p L p N D ) (long) J o = q ( D n W p N A + D p W n N D ) (short) Non-ideal diodes: I = I o (e q(v A IR s )/kt 1) J gen = q τ o W Photovoltaics: V oc = nkt ln q (J sc ) J o J PV = J o (e qv A/kT 1) J sc Small signal model: G d = I D+I o kt/q C J (V R ) = K sε o A K sεov bi qn A = A qksεona V bi C D = G d τ n MS diode properties: qv bi = Φ M Φ S Φ BP = χ + E G Φ M Φ BN = Φ M χ J D = J o (e qv A/kT 1) J o = A T e Φ B/kT A = 4πqm k B h 3 = 10 m m o A cm K MOS capacitors: W = K sε o φ s qn A cm ℇ s = qn Aφ s K s ε o V cm Q B = qn A W(φ s ) = qk s ε o N A φ s C cm V G = V FB + φ s + Δφ ox = V FB + φ s Q s(φ s ) C ox C ox = K o ε o /x o V FB = Φ ms /q Q F /C ox C = C ox / [1 + K ow(φ s ) K s x o ] V T = Q B (φ F )/C ox + φ F Q n = C ox (V G V T ) MOSFETs: I D = WQ n (y = 0) v y (y = 0) I D = W L μ nc ox (V GS V T )V DS I D = WC ox v sat (V GS V T ) W L Square Law (for V GS V T ): I D = { μ nc ox [(V GS V T )V DS V DS /], 0 V DS V GS V T W μ L nc ox (V GS V T ), V DS V GS V T
Bipolar transistors: (assuming NPN, short emitter, base, and collector) Ebers-Moll Equations: